Register to reply 
Infinite sequences and series.. help! 
Share this thread: 
#1
Dec712, 06:09 AM

P: 217

Hi I don't understand the logic in the picture i added.
They say that "that sum of the series = the limit of the sequence" The limit is 2/3 BUT the sum, Ʃ, must be 2*1/(3*1+5) + (2*2/(2*3+5) + 2*3/(2*3+5) ...+ Which is obviously much larger than 2/3 if all the terms are added together?? it's like adding 2/3+2/3+2/3 which is = 6/2 =2?? Is there something I'm misunderstanding here? 


#2
Dec712, 06:23 AM

P: 3,101

what is the a sub n term? the a sub n terms are added together to get the s sub n
the sequence referred to is the s sub n terms as defined by the 2*n/(3*n+5) you don't sum these. Instead you must evaluate the term as n approaches infinity. so that when n is very large the sequence terms become 2*n / 3*n and the n drops out to leave 2/3 


#3
Dec712, 06:45 AM

P: 217

Well if we have the series Ʃa_{n}= 1/2^{n} i can see that the sum would add up to one.
Ahhh So the limit of one Lim 1 = 1 = the sum?? Is that how I should understand it? 


#4
Dec712, 06:52 AM

P: 217

Infinite sequences and series.. help!
Ahh so the partial sequence of Sn adds up to 2n/(3n+5) as you said (just like 1/2^2 adds up to one). so if we add up all the terms of some imaginary sequence we would get the sum to be 2n/(3n+5), so the sum is the end result and NOT 2*1/(3*1+5) + (2*2/(2*3+5) + 2*3/(2*3+5) ...+ which is the result of adding up the sums.
So the limit is the the sum because the sum is 2n/(3n+5) and when n gets big it's 2/3. Is that correctly understood? And thank you :) 


#5
Dec712, 07:00 AM

P: 217

So is it correctly understood that the limit of a sequence 1/2^n is zero but the limit of the partial sums of that sequence is 1? That's the diffrence?



#6
Dec712, 08:02 AM

P: 3,101

Look at a simpler example: Ʃ(2n+1) = n^2 a_{n} = (2n+1) s_{n} = n^2 so the sequence of s_{n} is { 1, 4, 9, 16, 25 ... } so as n approaches infinity what is the last term of the sequence of s_{n} 


#7
Dec712, 10:07 AM

P: 78

don't look at it that way
divide each term by the highest power variable and take each portion of the equations limit, separately. a quick way of doing it is to take a ratio of coefficients of variables in this case of a_{n} 


#8
Dec712, 11:17 AM

P: 217

Hmm well Okay first, Ʃ(2n+1) means 2*1+1 +2*2+1 ...(2n+1) this is the action of adding up a sequence of numbers, right? Isn't that why we use a summation symbol? Isn't it the definition of summation "Sum" = add up?
Now follow me here (and please tell me it's correctly understood :) then I'll be happy ) The sum of this series 1/2+1/4+1/8...+ has a pattern 1/2^n and we denote it Ʃ(1/2^n ), By taking a partial sum Sn= 1/2+1/4+1/8 We check if this sum approaches a limit (in this case Lim Sn=1) So the sum of the series Ʃ(1/2^n ) is the limit of the sequence of partial sums, right?? The limit of 1/2^n approaches zero but the limit of the partial sum Sn approaches one. This must be correct, please :D 


#9
Dec712, 11:30 AM

P: 3,101

With respect to your prior statement, it would be better to say:
partial sum of Sn is given by 2n/(3n+5) and this approaches 2/3 as n approaches infinity. perhaps we're splitting hairs here but I didn't want you to think that the 2n/(3n+5) is the same as s1+s2+s3... 


#10
Dec712, 11:39 AM

P: 217

Thank you! :)
I see because you wanted me to understand that 2n/(3n+5) is the infinite series (when n goes to infinity) and s1+s2+s3... is a sequence of numbers. 


#11
Dec712, 12:51 PM

Mentor
P: 21,409

$$ \sum_{n = 1}^{\infty} \frac{2n}{3n + 5}$$ 2n/(3n + 5) is merely the nth term in the series or the nth term in the sequence that underlies the series. Every infinite series involves two sequences: 1. The sequence that makes up the terms in the series. 2. The sequence of partial sums. Using the first example above, the sequence of terms would be {2/8, 4/11, 6/14, ..., 2n/(3n + 5), ...} The seqence of partial sums would be {2/8, 2/8 + 4/11, 2/8 + 4/11 + 6/14, ...} (I haven't bothered to simplify the fractions or combine them.) A series converges if its sequence of partial sums converges. 


#12
Dec712, 01:09 PM

P: 217

S1 = a1 S2= a1 +a1 S3 = a1+a2+a3. So the nth term of a_{n} is 2n/(3n + 5) and the nth term of Sn as i goes from 1 to n gives approaches the limit/sum for the finite series if it converges. 


#13
Dec712, 01:13 PM

P: 217

But wait, this part: 2n/(3n + 5) is merely the nth term in the series or the nth term in the sequence that underlies the series.
What's the difference between the series and the sequence that underlies the series? The series is the infinite sequence of numbers right? and the sequence is a part of that series? 


#14
Dec712, 01:19 PM

P: 217

And to this "1. The sequence that makes up the terms in the series."
A term is 2/8 , so don't you mean "the terms that make up the sequence (a1 +a2+a3) in the series" 


#15
Dec712, 01:46 PM

Mentor
P: 21,409

The sequence is a list, albeit an infinitely long list: {a_{1}, a_{2}, a_{3} ,,, a_{n}...} The series (AKA infinite series) is the sum that you get when you add all of the terms in the sequence. Here S_{1} = a_{1} S_{2} = a_{1} + a_{2} = S_{1} + a_{2} S_{3} = a_{1} + a_{2} + a_{3}= S_{2} + a_{3} and so on. In general, S_{n} = ## \sum_{i = 1}^n a_n##, which is a finite sum. 


#16
Dec712, 02:11 PM

P: 217

I'm sory, You are right I'm confusing things together. I sat down, reread everything, and here it is. I understand it now, I really spent time thinking about it.
We have a sequence of number fx the sequence 1/2^{n}, and we can test the limit of such a sequence, by doing this we want to find out what the sequence approaches as n → ∞ Lim1/2^{n} = 0 If we add the terms of an infinite sequence we get what we call an Infinite series. The series is a number/sum denoted Ʃ(a^{n}) consisting of the addition of the numbers in the sequence a^{n}. Instead of adding the whole sequence to infinity, we add a part of the sequence. We call this The partial Sums, denoted Sn. Sn = A1+a2+a3…an(…=infinity) We can test if this series of numbers converges toward a finite number, or if it diverges to infinity by using partial sum. Example If we have a sequence a^{n} = 1/2^{n} The sequence is: 1/2 ,1/4, 1/8…,1/n^{n}...(...= going to infinity) The limit of the sequence if we let n go toward infinity is Lim 1/2^{n} =0 On the other hand The series Ʃ(1/2^{n}) also has a limit. We can prove this by a rule that says if the Sequece of the partial sum {Sn} is convergent (which it is, because the limit of the sequence {Sn}=1/2 ,1/2+1/4,1/2+1/4+ 1/8…,1/n^{n} is 1} then the series Ʃ(1/2^{n}) is convergent, and it is because it converges toward 1. 


#17
Dec712, 02:22 PM

P: 217

You were right, I was confusing the sequence of terms with the sequence of partial sums, that was what confused me so much in my calc book. Thank you! :)



#18
Dec712, 03:23 PM

Mentor
P: 21,409

There are a couple of nits below, but otherwise you seem to have a good handle on the ideas. S_{n} = a_{1} + a_{2} + ... + a_{n}. However, as n increases, the number of terms that make up S_{n} gets larger, but there are always n terms. A little work shows that the sequence of partial sums can be written as {1/2, 3/4, 7/8, ... , (2^{n}  1)/2^{2}, ...} Good work! 


Register to reply 
Related Discussions  
Infinite Series and Sequences  Science & Math Textbooks  1  
Infinite sequences and series  conv or div  sigma(e^(1/n)/n)  Calculus & Beyond Homework  8  
Power of a set of infinite sequences  Set Theory, Logic, Probability, Statistics  0  
Divergent Harmonic Series, Convergent PSeries (Cauchy sequences)  Calculus & Beyond Homework  1  
Infinite series/sequences  Calculus  4 